JEE Physics 2025 Question with Solution

Given: Two slabs of materials $1$ and $2$ have equal side $l$, thicknesses $d_1$ and $d_2$ with $d_2 = 2d_1$. Equal shearing force $F$ is applied on the narrow faces, and the deformation angles satisfy $\theta_2 = 2\theta_1$. Shear modulus of material $1$ is $\eta_1 = 4 \times 10^9 \, \text{N/m}^2$.

Find: The value of $x$ in $\eta_2 = x \times 10^9 \, \text{N/m}^2$.

Using the relation from the solution working:

$$2\theta_1 = \theta_2$$ $$2 \frac{\sigma_1}{\eta_1} = \frac{\sigma_2}{\eta_2}$$

For equal shearing force on narrow faces, the shear stresses are

$$\sigma_1 = \frac{F}{l d_1}, \qquad \sigma_2 = \frac{F}{l d_2}$$

Substituting these into the strain relation,

$$2\left(\frac{F}{l d_1 \eta_1}\right) = \frac{F}{l d_2 \eta_2}$$

Given $d_2 = 2d_1$,

$$2\left(\frac{F}{l d_1 \eta_1}\right) = \frac{F}{2l d_1 \eta_2}$$

Cancelling common factors,

$$\eta_2 = \frac{\eta_1}{4}$$

Now substitute $\eta_1 = 4 \times 10^9 \, \text{N/m}^2$:

$$\eta_2 = 1 \times 10^9 \, \text{N/m}^2$$

Hence,

$$x = 1$$

Therefore, the value of $x$ is $1$.

Given: $\theta_2 = 2\theta_1$, $d_2 = 2d_1$, and $G_1 = 4 \times 10^9 \, \text{N/m}^2$.

Find: $G_2 = x \times 10^9 \, \text{N/m}^2$.

For shear deformation,

$$G = \frac{\text{shear stress}}{\text{shear strain}}$$

so

$$\theta = \frac{\text{shear stress}}{G}$$

Since the force is the same but the narrow face areas are different,

$$\tau_1 = \frac{F}{l d_1}, \qquad \tau_2 = \frac{F}{l d_2}$$

Therefore,

$$\theta_1 = \frac{F}{l d_1 G_1}, \qquad \theta_2 = \frac{F}{l d_2 G_2}$$

Using $\theta_2 = 2\theta_1$,

$$\frac{F}{l d_2 G_2} = 2\left(\frac{F}{l d_1 G_1}\right)$$

Substitute $d_2 = 2d_1$:

$$\frac{F}{2l d_1 G_2} = \frac{2F}{l d_1 G_1}$$

Cancelling $\frac{F}{l d_1}$,

$$\frac{1}{2G_2} = \frac{2}{G_1}$$ $$G_2 = \frac{G_1}{4}$$

Now,

$$G_2 = \frac{4 \times 10^9}{4} = 1 \times 10^9 \, \text{N/m}^2$$

So the required value is $x = 1$.